A chemical amount changes at rate . Which is an antiderivative of ?
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AP Calculus AB Quiz
Practice Finding Antiderivatives And Indefinite Integrals in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A chemical amount changes at rate A′(t)=21t4−3t2. Which is an antiderivative of A′(t)?
This quiz focuses on Finding Antiderivatives And Indefinite Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
A chemical amount changes at rate A′(t)=21t4−3t2. Which is an antiderivative of A′(t)?
Explanation: This problem involves finding an antiderivative through fundamental integration principles. To find an antiderivative of A'(t) = (1/2)t⁴ - 3t², we apply the power rule: ∫tⁿ dt = t^(n+1)/(n+1) + C. For (1/2)t⁴, we get (1/2)·t⁵/5 = (1/10)t⁵, and for -3t², we get -3·t³/3 = -t³. Thus, the antiderivative is (1/10)t⁵ - t³ + C. Choice D ((1/10)t⁵ - 3t² + C) incorrectly integrates the second term, forgetting to increase the exponent—a mistake showing confusion about the integration process. Always remember: integration reverses differentiation by increasing powers and dividing by the new power.
A chemical amount changes at rate A′(t)=4t9+11. Which expression is a general antiderivative of A′(t)?
Explanation: This problem requires finding a general antiderivative of A'(t) = 4t⁹ + 11, demonstrating basic antiderivative reasoning. To find the antiderivative, we apply the power rule for integration: ∫tⁿdt = t^(n+1)/(n+1) + C. For 4t⁹, we get 4t¹⁰/10 = (2/5)t¹⁰; for the constant 11, we get 11t. Combining these terms with the constant of integration gives (2/5)t¹⁰ + 11t + C. Choice B ((4/10)t¹⁰ + 11t + C) shows the unsimplified fraction 4/10 instead of the reduced form 2/5, though both are mathematically equivalent. The key integration strategy is to apply the power rule carefully and simplify fractions when possible for cleaner expressions.
The marginal cost is M(x)=12x5+3x2. Which is an antiderivative of M(x)?
Explanation: This problem requires finding an antiderivative by applying the power rule for integration to each term. For M(x) = 12x⁵ + 3x², we integrate term by term: 12x⁵ becomes 12x⁶/6 = 2x⁶, and 3x² becomes 3x³/3 = x³. The antiderivative is 2x⁶ + x³ + C. Choice B (2x⁶ + 3x³ + C) incorrectly keeps the coefficient 3 when integrating 3x², failing to divide by the new exponent 3. Remember the integration formula: ∫axⁿ dx = axⁿ⁺¹/(n+1) + C, where you must divide by the new exponent to reverse the chain rule effect from differentiation.
The derivative of a function is f′(x)=21x7−6x. Which is a general antiderivative of f′(x)?
Explanation: This problem asks for a general antiderivative of f'(x) = (1/2)x⁷ - 6x, testing basic antiderivative reasoning skills. To find the antiderivative, we apply the power rule for integration to each term separately. For (1/2)x⁷, we get (1/2)·x⁸/8 = x⁸/16; for -6x, we get -6x²/2 = -3x². Including the constant of integration yields (1/16)x⁸ - 3x² + C. Choice B ((1/16)x⁸ - 6x² + C) incorrectly integrates -6x as -6x² instead of -3x², failing to divide by the new exponent 2. The transferable strategy is to remember that when integrating axⁿ, the result is a·x^(n+1)/(n+1), requiring division by the new exponent.
A particle’s velocity is v(t)=6t2−4t+9. Which function is an antiderivative of v(t)?
Explanation: This problem requires finding an antiderivative using basic integration rules. To find an antiderivative of v(t) = 6t² - 4t + 9, we reverse the differentiation process by applying the power rule for integration: ∫tⁿ dt = tⁿ⁺¹/(n+1) + C. For 6t², we get 6t³/3 = 2t³; for -4t, we get -4t²/2 = -2t²; and for 9, we get 9t. Combining these terms gives 2t³ - 2t² + 9t + C. Choice E (2t³ - 2t² + 9 + C) incorrectly integrates the constant 9 as 9 instead of 9t, forgetting that the integral of a constant is that constant times the variable. Remember: to verify an antiderivative, differentiate your answer and check if you get back the original function.
A function has derivative g′(x)=3x8−10x4+2. Which is a general antiderivative of g′(x)?
Explanation: This problem asks for a general antiderivative of g'(x) = 3x⁸ - 10x⁴ + 2, testing basic antiderivative reasoning skills. To find the antiderivative, we apply the power rule for integration to each term. For 3x⁸, we get 3x⁹/9 = (1/3)x⁹; for -10x⁴, we get -10x⁵/5 = -2x⁵; for the constant 2, we get 2x. Combining these with the constant of integration yields (1/3)x⁹ - 2x⁵ + 2x + C. Choice D ((1/9)x⁹ - 2x⁵ + 2x + C) incorrectly integrates 3x⁸ as (1/9)x⁹ instead of (1/3)x⁹, making an error in simplifying 3/9. The transferable strategy is to carefully simplify coefficients after applying the power rule: when integrating ax^n, the result is a/(n+1)·x^(n+1).
A runner’s velocity is v(t)=23t4−5t. Which is an antiderivative of v(t)?
Explanation: Finding antiderivatives requires careful application of the power rule, especially with fractional coefficients. To integrate v(t) = (3/2)t⁴ - 5t, we apply the rule to each term: (3/2)t⁴ becomes (3/2)t⁵/5 = (3/10)t⁵, and -5t becomes -5t²/2 = -(5/2)t². The antiderivative is (3/10)t⁵ - (5/2)t² + C. Choice E ((3/2)t⁵ - (5/2)t² + C) incorrectly keeps the original coefficient 3/2 when integrating (3/2)t⁴, forgetting to divide by the new exponent 5. When integrating terms with fractions, multiply the coefficient by 1/(n+1), which often creates more complex fractions that must be simplified.
The marginal cost is C′(x)=9x2+7x6. Which is a general antiderivative for C′(x)?
Explanation: This problem asks for a general antiderivative of the marginal cost function C'(x) = 9x² + 7x⁶, testing basic antiderivative reasoning skills. To find the antiderivative, we apply the power rule for integration: for xⁿ, the antiderivative is x^(n+1)/(n+1). For 9x², we get 9x³/3 = 3x³; for 7x⁶, we get 7x⁷/7 = x⁷. Including the constant of integration, the general antiderivative is 3x³ + x⁷ + C. Choice D (3x³ + x⁷) omits the crucial constant of integration C, which represents all possible vertical translations of the antiderivative function. The transferable strategy is to always include +C in indefinite integrals, as this constant captures the infinite family of functions that all have the same derivative.
A population grows at rate P′(t)=12t5−5t2. Find an antiderivative of P′(t).
Explanation: This problem requires finding an antiderivative of the population growth rate P'(t) = 12t⁵ - 5t², demonstrating basic antiderivative reasoning. To find the antiderivative, we reverse the differentiation process using the power rule for integration. For 12t⁵, we get 12t⁶/6 = 2t⁶; for -5t², we get -5t³/3 = -5/3·t³. Combining these with the constant of integration gives 2t⁶ - (5/3)t³ + C. Choice B (2t⁶ - 5t³ + C) incorrectly integrates -5t² as -5t³ instead of -(5/3)t³, forgetting to divide by the new exponent. The key integration strategy is to carefully apply the power rule formula ∫tⁿdt = t^(n+1)/(n+1) + C, ensuring you divide by the new exponent (n+1).
A machine’s power usage rate is p(t)=7t2+14t5. Find ∫p(t)dt.
Explanation: This problem requires finding ∫p(t)dt where p(t) = 7t² + 14t⁵, demonstrating basic antiderivative reasoning through integration notation. To find the indefinite integral, we apply the power rule to each term: ∫tⁿdt = t^(n+1)/(n+1) + C. For 7t², we get 7t³/3 = (7/3)t³; for 14t⁵, we get 14t⁶/6 = (14/6)t⁶ = (7/3)t⁶. Including the constant of integration gives (7/3)t³ + (7/3)t⁶ + C. Choice B shows (14/6)t⁶ instead of the simplified (7/3)t⁶, though both are equivalent since 14/6 = 7/3. The key strategy is to simplify fractions in your final answer and always include +C for indefinite integrals.
The slope of a curve is dxdy=15x4−2x3. Which y is an antiderivative?
Explanation: This problem asks for an antiderivative y given that dy/dx = 15x⁴ - 2x³, testing basic antiderivative reasoning skills. To find y, we integrate the derivative using the power rule for integration. For 15x⁴, we get 15x⁵/5 = 3x⁵; for -2x³, we get -2x⁴/4 = -(1/2)x⁴. Including the constant of integration yields y = 3x⁵ - (1/2)x⁴ + C. Choice B (3x⁵ - 2x⁴ + C) incorrectly integrates -2x³ as -2x⁴ instead of -(1/2)x⁴, forgetting to divide by the new exponent 4. The transferable integration-differentiation reversal strategy is to always divide by (n+1) when integrating xⁿ, which reverses the multiplication by n that occurs during differentiation.
A particle’s velocity is v(t)=6t2−4t+9. Which function is an antiderivative of v(t)?
Explanation: This problem requires finding an antiderivative of the velocity function v(t) = 6t² - 4t + 9, which tests basic antiderivative reasoning skills. To find an antiderivative, we reverse the differentiation process by applying the power rule in reverse: for each term tⁿ, the antiderivative is t^(n+1)/(n+1). For 6t², we get 6t³/3 = 2t³; for -4t, we get -4t²/2 = -2t²; and for the constant 9, we get 9t. Since antiderivatives are not unique, we must include the constant of integration C, giving us 2t³ - 2t² + 9t + C. Choice A (2t³ - 2t² + 9t) is tempting but incorrect because it omits the essential constant of integration that represents the family of all antiderivatives. The key strategy is to remember that indefinite integrals always include +C to represent all possible vertical shifts of the antiderivative function.
A particle’s velocity is v(x)=27x5−3x2. Which is an antiderivative of v(x)?
Explanation: This problem requires finding an antiderivative by applying the power rule for integration to each term. For v(x) = 7/2·x⁵ - 3x², we integrate term by term: the antiderivative of 7/2·x⁵ is (7/2)·x⁶/6 = 7x⁶/12, and the antiderivative of -3x² is -3x³/3 = -x³. The complete antiderivative is 7/12·x⁶ - x³ + C. Choice D (7/12·x⁶ - x³) is tempting but incorrect because it lacks the constant of integration C, which represents the infinite family of functions that all have the same derivative. The strategy for finding antiderivatives is to reverse differentiation: increase the power by 1, divide by the new power, and always include the constant C.
The acceleration of a cart is a(t)=8t3+5. Which function is an antiderivative of a(t)?
Explanation: Finding an antiderivative requires reversing the differentiation process using integration rules. For a(t) = 8t³ + 5, we apply the power rule: the antiderivative of 8t³ is 8t⁴/4 = 2t⁴, and the antiderivative of 5 is 5t. Therefore, the complete antiderivative is 2t⁴ + 5t + C, where C is the constant of integration. Choice B (2t⁴ + 5t) is incorrect because it omits the constant C, which is essential for representing all possible antiderivatives that differ by a constant. When integrating, remember to add 1 to the exponent and divide by the new exponent, and always include the constant C unless working with definite integrals.
A car’s acceleration is a(t)=12t−7. Which is an antiderivative of a(t)?
Explanation: This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of a(t) = 12t - 7, increase the exponent of each term by one and divide by the new exponent, resulting in (12/2)t² - 7t + C. This process reverses differentiation because differentiating 6t² - 7t + C yields 12t - 7, matching the original acceleration function. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice B fails because it omits the +C, which is necessary for the general antiderivative. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
A population changes at rate p(t)=7−9t2. Which is an antiderivative of p(t)?
Explanation: This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of p(t) = 7 - 9t², increase the exponent of each term by one and divide by the new exponent, resulting in 7t - (9/3)t³ + C. This process reverses differentiation because differentiating 7t - 3t³ + C yields 7 - 9t², matching the original population change rate. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice B fails because it incorrectly uses a coefficient of -9 instead of -3 for the t³ term. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
A particle’s velocity is v(t)=6t2−4t+9. Which is an antiderivative of v(t)?
Explanation: This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of v(t) = 6t² - 4t + 9, increase the exponent of each term by one and divide by the new exponent, resulting in (6/3)t³ - (4/2)t² + 9t + C. This process reverses differentiation because differentiating 2t³ - 2t² + 9t + C yields 6t² - 4t + 9, matching the original velocity function. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice A fails because it omits the +C, which is necessary for the general antiderivative. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
The velocity of a runner is v(t)=15t2+6t−1. Which is an antiderivative of v(t)?
Explanation: This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of v(t) = 15t² + 6t - 1, increase the exponent of each term by one and divide by the new exponent, resulting in (15/3)t³ + (6/2)t² - t + C. This process reverses differentiation because differentiating 5t³ + 3t² - t + C yields 15t² + 6t - 1, matching the original velocity. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice D fails because it incorrectly uses a coefficient of 15 instead of 5 for the t³ term. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
The rate of change is k′(t)=13t12. Which is an antiderivative of k′(t)?
Explanation: This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of k'(t) = 13t¹², increase the exponent by one and divide by the new exponent, resulting in (13/13)t¹³ + C. This process reverses differentiation because differentiating t¹³ + C yields 13t¹², matching the original rate of change. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice B fails because it incorrectly uses a coefficient of 13 instead of 1 for the t¹³ term. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
The derivative is m′(x)=4x3+9x8. Which is an antiderivative of m′(x)?
Explanation: This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of m'(x) = 4x³ + 9x⁸, increase the exponent of each term by one and divide by the new exponent, resulting in (4/4)x⁴ + (9/9)x⁹ + C. This process reverses differentiation because differentiating x⁴ + x⁹ + C yields 4x³ + 9x⁸, matching the original derivative. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice C fails because it incorrectly uses coefficients of 4 and 9 without dividing properly. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.